what mass of lead(II)nitrate can be formed from reaction of 25.0ml of .100M sodium sulfate and 40.0ml of .200M lead(II)nitrate solutions?

I don't think you posted the question you want; i.e., I think you mis-read the question. It makes no sense as is.

To determine the mass of lead(II) nitrate formed, we first need to calculate the moles of sodium sulfate and lead(II) nitrate using their respective molarity and volume.

Step 1: Calculate the moles of sodium sulfate:
Moles = Molarity × Volume (in liters)
Moles of sodium sulfate = 0.100 M × 0.0250 L = 0.00250 moles

Step 2: Calculate the moles of lead(II) nitrate:
Moles of lead(II) nitrate = 0.200 M × 0.0400 L = 0.00800 moles

Step 3: Determine the limiting reagent:
To determine the limiting reagent, compare the moles of both reactants. The reactant with fewer moles will be the limiting reagent, as it will be fully consumed in the reaction.

Moles of sodium sulfate: 0.00250 moles
Moles of lead(II) nitrate: 0.00800 moles

Since sodium sulfate has fewer moles, it is the limiting reagent.

Step 4: Use stoichiometry to find the moles of lead(II) nitrate:
From the balanced chemical equation, we know that one mole of sodium sulfate reacts with one mole of lead(II) nitrate, forming one mole of lead(II) nitrate.

Using the ratio of 1:1, the moles of lead(II) nitrate formed will be 0.00250 moles.

Step 5: Calculate the mass of lead(II) nitrate:
Mass = Moles × Molar mass
Molar mass of lead(II) nitrate (Pb(NO3)2) = 331.21 g/mol

Mass of lead(II) nitrate = 0.00250 moles × 331.21 g/mol = 0.828 g

Therefore, the mass of lead(II) nitrate that can be formed is 0.828 grams.